Surface Areas And Volumes Worksheets With Answers - Class 10 - 2012
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1
22012
Maths Class 10 Chapter 13
Exercise 13.1
Surface Areas and Volumes
Q3: A toy is in the form of a cone of radius 3.5 cm mounted
Summary
on a hemisphere of same radius. The total height of the toy
In this chapter, you have studied the following points:
22
is 15.5 cm. Find the total surface area of the toy.(Use �� =
1. To determine the surface area of an object formed by
)
7
combining any two of the basic solids, namely, cuboid, cone,
Ans:
cylinder, sphere and hemisphere.
2. To find the volume of objects formed by combining any
two of a cuboid, cone, cylinder, sphere and hemisphere.
3. Given a right circular cone, which is sliced through by a
plane parallel to its base, when the smaller conical portion is
removed, the resulting solid is called a Frustum of a Right
Circular Cone.
4. The formulae involving the frustum of a cone are:
1
2
2
(i) ������������ ���� �� �������������� ���� �� �������� = =
���� ��
+ ��
+ ��
��
1
2
1
2
3
It can be observed that the radius of the conical part and the
(ii) ������������ �������������� �������� ���� �� �������������� ���� �� �������� = ����(��1 + ��2)
2
hemispherical part is same (i.e., 3.5 cm).
where �� = ��
+ (��
− ��
)
1
2
2
2
(iii) ���������� �������������� �������� ���� �� �������������� ���� �� �������� = ���� ��1 + ��2 + �� ��
+ ��
Height of hemispherical part = Radius (r) = 3.5 cm
1
2
Where,
Height of conical part (h) = 15.5 −3.5 = 12 cm
h = vertical height of the frustum,
2
2
���������� ������������ �� ���� �������������� �������� = ��
+ ��
l = slant height of the frustum
��1 and ��2 are radii of the two bases (ends) of the frustum.
2
2
= (3.5)
+ (12)
2
7
NCERT Solutions
Exercise 13.1
Q & A
=
+ 144
2
NCERT Book - Page No 244
3
49+576
625
25
Q1:2 cubes each of volume 64 cm
are joined end to end.
=
=
=
����
4
4
2
Find the surface area of the resulting cuboids.
Total surface area of toy = CSA of conical part + CSA of hemispherical part
Ans:
2
= 2������ + 2����
Given that,
22
7
25
22
7
7
=
×
×
+ 2 ×
×
×
3
Volume of cubes = 64 cm
7
2
2
7
2
2
3
2
= 137.5 + 77 = 214.5 ����
(Edge)
= 64
Q4: A cubical block of side 7 cm is surmounted by a
Edge = 4 cm
hemisphere. What is the greatest diameter the hemisphere
22
can have? Find the surface area of the solid. (Use �� =
)
7
Ans:
If cubes are joined end to end, the dimensions of the
resulting cuboid will be 4 cm, 4 cm, 8 cm.
∴ �������������� �������� ���� �������������� = 2 ���� + ���� + ����
= 2 4 × 4 + 4 × 8 + 4 × 8
= 2 16 + 32 + 32
From the figure, it can be observed that the greatest
2
diameter possible for such hemisphere is equal to the cube’s
= 2 16 + 64 = 2 × 80 = 160 ����
Q2: A vessel is in the form of a hollow hemisphere mounted
edge, i.e., 7cm.
7
by a hollow cylinder. The diameter of the hemisphere is 14
Radius (r) of hemispherical part =
= 3.5cm
2
cm and the total height of the vessel is 13 cm. Find the inner
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part
22
− Area of base of hemispherical part
surface area of the vessel. (Use �� =
)
2
2
2
7
= 6 ��������
+ 2����
− ����
Ans:
2
2
= 6 ��������
+ ����
22
7
7
2
∴ ���������� �������������� �������� ���� ���������� = 6 7
+
×
×
7
2
2
2
= 294 + 38.5 = 332.5 ����
Q5: A hemispherical depression is cut out from one face of a
cubical wooden block such that the diameter l of the
hemisphere is equal to the edge of the cube. Determine the
surface area of the remaining solid.
Ans:
Diameter of hemisphere = Edge of cube = l
It can be observed that radius (r) of the cylindrical part and
��
Radius of hemisphere =
the hemispherical part is the same (i.e., 7 cm).
2
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part
Height of hemispherical part = Radius = 7 cm
− Area of base of hemispherical part
Height of cylindrical part (h) = 13 −7 = 6 cm
2
2
2
= 6 ��������
+ 2����
− ����
Inner surface area of the vessel = CSA of cylindrical part +
2
2
= 6 ��������
+ ����
CSA of hemispherical part
2
= 2������ + 2����
2
��
2
22
22
∴ ���������� �������������� �������� ���� ���������� = 6 ��
+ �� ×
Inner surface area of vessel = 2 ×
× 7 × 6 + 2 ×
× 7 × 7
2
7
7
2
����
1
2
2
2
= 44 6 + 7 = 44 × 13
= 6��
+
=
24 + �� ��
��������
4
4
2
= 572 ����
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